Is the function given below continuous/differentiable at $x=1$ ? $g(x)=\begin{cases} x^2+2x&,&x\leq1 \\\\ 4x-1&,&x>1 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Checking for continuity at $x=1$ For the function to be continuous at $x=1$, we need the two-sided limit $\lim_{x\to 1}g(x)$ to exist and be equal to $g(1)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 1^-}g(x)$ and $\lim_{x\to 1^+}g(x)$ exist and are equal to $g(1)$. According to $g$ 's definition, $g(1)=(1)^2+2(1)=3$. $\lim_{x\to 1^-}g(x)$ $x^2+2x$ evaluated at $x=1$ is equal to $3$. Since $x^2+2x$ is continuous, we can be certain that $\lim_{x\to 1^-}g(x)=3$. $\lim_{x\to 1^+}g(x)$ $4x-1$ evaluated at $x=1$ is equal to $3$. Since $4x-1$ is continuous, we can be certain that $\lim_{x\to 1^+}g(x)=3$. We saw that the two one-sided limits exist and are equal to $g(1)$, so the function is continuous at $x=1$. Checking for differentiability at $x=1$ For the function to be differentiable at $x=1$, we need the two-sided limit $\lim_{x\to 1}\dfrac{g(x)-g(1)}{x-1}=\,\lim_{x\to 1}\dfrac{g(x)-3}{x-1}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 1^-}\dfrac{g(x)-3}{x-1}$ and $\lim_{x\to 1^+}\dfrac{g(x)-3}{x-1}$ exist and have the same value. $\lim_{x\to 1^-}\dfrac{g(x)-3}{x-1}=4$ $\lim_{x\to 1^+}\dfrac{g(x)-3}{x-1}=4$ We saw that the two one-sided limits exist and are equal, so the function is differentiable at $x=1$. In conclusion, the function is both continuous and differentiable at $x=1$.